Please help

**Jon** · March 12th 2006, 10:20 AM

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**ThePerfectHacker** · March 12th 2006, 04:12 PM

Originally Posted by Jon

need help ASAP Six dwarfs: Sleepy, Happy, Bashful, Doc, Grumpy, and Wishbone are available for a behavior experiment. In the experiment, 4 of therm are selcted and arranged in a row. Find the probability for the following events:

Doc is first
Wishbone is last
Bashful is first and Happy is last
Happy is first, Grumpy is second , Doc is thrid and Wishbone is last
Sleepy is first, second or third but not last
Wishbone is not selected

Doc is first:
He needs to be selected AND he would be the first. The probability that he is selected is $\frac{_5C_3\cdot _1C_1}{_6C_4}=\frac{2}{3}$
The probability that he is first is $1/4$ thus, the answer is $\frac{1}{4}\cdot \frac{2}{3}=\frac{1}{6}$

Wishbone:
Same thing as in the first problem, same answer.

Bashful is first, Happy is last:First they need to be selected out of 6 people which is $\frac{_2C_2\cdot _4C_2}{_6C_4}=\frac{2}{5}$
Now bashful is first which is $1/4$ and hapy is last which is $1/3$ because only 3 poeple remain after bashful. Thus,
$\frac{2}{5}\cdot \frac{1}{4}\cdot \frac{1}{3}=\frac{1}{30}$

For the next problem, you need to select all 4 which is $\frac{_4C_4}{_6C_4}=\frac{1}{15}$ , Now you need them is proper order which is for the first one $1/4$ for the second $1/3$ for the third $1/2$ and for the last $1/1$ Thus,
$\frac{1}{15}\cdot \frac{1}{4}\cdot \frac{1}{3}\cdot \frac{1}{2}\cdot \frac{1}{1}=\frac{1}{360}$

Sleepy Not last: Add that he is first or second or third. Which is $\frac{1}{6}$ as in the first problem. Thus, $\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$

Wishbone not selected: If the probability of selecting him is $\frac{2}{3}$ then the probability of not selecting him is $\frac{1}{3}$
Hope this helps, and also that I did not erred.

**Jon** · March 12th 2006, 04:18 PM

I did come up with some of the answers, but on others I see where I made my mistake.

Thanks.

**ThePerfectHacker** · March 12th 2006, 06:10 PM

Originally Posted by Jon

I did come up with some of the answers, but on others I see where I made my mistake.

Thanks.

Welcome, no need to delete the post let other members see if there are any other ways of answering it or maybe they want to see the answer.

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